Question: Divide the following complex numbers. $\dfrac{10+40i}{-3+5i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${-3-5i}$. $ \dfrac{10+40i}{-3+5i} = \dfrac{10+40i}{-3+5i} \cdot \dfrac{{-3-5i}}{{-3-5i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(10+40i) \cdot (-3-5i)} {(-3)^2 - (5i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(10+40i) \cdot (-3-5i)} {(-3)^2 - (5i)^2} $ $ = \dfrac{(10+40i) \cdot (-3-5i)} {9 + 25} $ $ = \dfrac{(10+40i) \cdot (-3-5i)} {34} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({10+40i}) \cdot ({-3-5i})} {34} $ $ = \dfrac{{10} \cdot {(-3)} + {40} \cdot {(-3) i} + {10} \cdot {-5 i} + {40} \cdot {-5 i^2}} {34} $ $ = \dfrac{-30 - 120i - 50i - 200 i^2} {34} $ Finally, simplify the fraction. $ \dfrac{-30 - 120i - 50i + 200} {34} = \dfrac{170 - 170i} {34} = 5-5i $